Does there exist some subring $k \subset \mathbb{C}$ such that the following assertion holds?

  • ($k$-Hodge conjecture) For each nonsingular algebraic variety $X$ over $\mathbb{C}$, and each $q = 0, 1, \ldots, \dim_{\mathbb{C}}(X)$, each class $\mathfrak{z} \in H^{2q}(X; k) \cap H^{q,q}(X)$ is a $k$-linear combination of classes of algebraic cycles.

Atiyah-Hirzebruch proved in 1961 that $k \neq \mathbb{Z}$. As commented by Ben Wieland below, the argument of Atiyah-Hirzebruch also shows that $k \neq \mathbb{Z}_{(p)}$ for any prime $p$.

Remark: If the millenium problem is true, then we may take $k = \mathbb{Q}$.

(The above post has been slightly edited from the original).

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    $\begingroup$ $\mathbb Q$ is sharp in both directions. The Atiyah-Hirzebruch argument works over $\mathbb Z_{(p)}$, which is as large as you can get. In the other direction, the $k$- version does not imply the $K$-version. On the contrary, the $\mathbb C$-version is false, see the product of any pair of non-isogenous elliptic curves. $\endgroup$ Dec 1, 2016 at 2:24
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    $\begingroup$ Can we conclude that the $k$-Hodge conjecture false for all $k \neq \mathbb{Q}$? $\endgroup$
    – user94803
    Dec 1, 2016 at 5:31
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    $\begingroup$ Related to this, you have the following theorem due to C. Voisin. Assume that the Hodge conjecture is known for varieties $X_{\bar{\mathbb{Q}}}$ defined over $\bar{\mathbb{Q}}$ and (weakly) absolute Hodge classes $\alpha$ on them. Then the $\mathbb{Q}$-Hodge conjecture is true for (weakly) absolute Hodge classes. This is Proposition 0.2 in "Hodge loci and absolute Hodge classes". $\endgroup$
    – diverietti
    Dec 2, 2016 at 12:19

1 Answer 1


The $k$-Hodge conjecture is false for any $k$ containing an irrational real $\alpha$.

Indeed we may clearly assume $\alpha$ is negative. Consider the elliptic curve $E = \mathbb C / \langle 1, \sqrt{\alpha} \rangle$. Let $e_1$ and $e_2$ be a basis of $H^1(E,\mathbb Z)$ such that the map to $H^{1,0}(E)$ sends $e_1$ to $1$ and $e_2$ to $\sqrt{\alpha}$. Then in $H^1(E,k) \otimes H^1(E,k) \subset H^2(E \times E, k)$, consider the class $$\alpha( e_1 \otimes e_1) - (e_2 \otimes e_2)$$

In the natural projection to $H^{2,0}$, it is sent to $\alpha(1) - (\sqrt{\alpha})^2=0$. In the natural projection to $H^{2,0}$, which is the complex conjugate of that, it is sent to $\alpha (1) - (-\sqrt{\alpha})^2=0$. So it lies in $H^{1,1}(E\times E) $.

But because $E$ is not CM (because $\alpha$ is irrational $\sqrt{\alpha}$ is not a quadratic irrational), the only algebraic cycle that contributes to $H^1(E) \otimes H^1(E)$ is the diagonal, which has class $e_1 \otimes e_2 - e_2 \otimes e_1$, and this class is not a multiple of it.

On the other hand, assuming the $\mathbb Q$-Hodge conjecture, we obtain the $k$-Hodge conjecture for any imaginary quadratic field $k$. Indeed, for $k= \mathbb Q(\sqrt{-D})$, $H^{2q}(X,\mathbb Q(\sqrt{-D}))= H^{2q}(X,\mathbb Q) + \sqrt{-D}H^{2q}(X,\mathbb Q)$. Because complex conjugation exchanges $H^{q,p}$ and $H^{p,q}$, it fixes $H^{q,q}$, so if a class in $H^{2q}(X,\mathbb Q(\sqrt{-D}))$ lands in $H^{q,q}$, both its real and imaginary parts do as well, so under the Hodge conjecture they are $\mathbb Q$-linear combinations of algebraic cycles, and the original is clearly a $\mathbb Q(\sqrt{-D})$-linear combination of algebraic cycles.

I'm not sure about number fields of degree at least $3$ that contain no real irrationals.

  • $\begingroup$ Will, thank you for your answer! Do you have a take on the case where $k$ is a subring of $\mathbb{Q}$? $\endgroup$
    – user94803
    Dec 5, 2016 at 0:56
  • $\begingroup$ @ColinTan Didn't Ben Wieland answer this? $\endgroup$
    – Will Sawin
    Dec 5, 2016 at 8:21
  • $\begingroup$ Does the $k$-Hodge conjecture being false for all $\mathbb{Z}_{(p)}$ imply that the $k$-Hodge conjecture is false whenever $k$ is a subring of $\mathbb{Q}$? $\endgroup$
    – user94803
    Dec 15, 2016 at 13:18
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    $\begingroup$ @ColinTan Yes, because every subring of $\mathbb Q$ is contained in $\mathbb Z_{(p)}$ for some $p$. $\endgroup$
    – Will Sawin
    Dec 15, 2016 at 14:14

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